# Simple Change of Variables in Lebesgue Integration

I'm not sure if these details matter, but anyway for this particular case, consider a compact abelian group $$G$$ with operation $$\cdot$$, a Haar measure $$\mu$$ on it and $$f$$ a non-trivial character on $$G$$. (I'm looking at this question: The integral of a character is $0$)

I've seen stated that for $$x,y\in G$$,

$$\int_G f(x)\>d\mu(x)=\int_G f(x\cdot y)\>d\mu(x\cdot y).$$

Is this true? If so why? What's the general theory behind it?

• Hint: recall the definition of (right/left) Haar measure (in particular, recall that $\mu(xA)=\mu(A)$). First show that the equation holds for simple functions, and then use a density argument to prove the general case. – Manuel Norman May 24 at 19:59
• @ManuelNorman I think the part that's tripping me up is if we know $\mu(xA)=\mu(A)$, why does that hold for $d\mu$? – J.Smith May 24 at 20:06
• This is usually a choice of notation: for instance, you could also simply write $d \mu$, or $\mu (dx)$; sometimes, it is not even explicitely written. – Manuel Norman May 24 at 20:12